-1
$\begingroup$

I have the following question at hand and I have to admit that I am not used to integral remainder form of taylor approximation. I am still trying to work around, so a couple of hints would be useful before a full answer.

Basically I have to determine the constant $c$ such that:

$\int\limits_{0}^{x}f(t)dt=\int\limits_{x}^{1}t^2f(t)dt + \frac{x^8}{8}+\frac{x^6}{6}-\frac{c}{24}$

Thanks in advance!

EDIT: I am sorry, I forgot to mention that:

$f:[0,1]\to\mathbb R$

I suppose thats important. And f is continuous, although I would imagine that is implicit...

EDIT_2:
Now, I would like to know if there is any other trickier solution or any approach using the integral remainder form of taylor expansion. Thanks!

$\endgroup$
  • $\begingroup$ Have you been told what $f(t)$ is or are you suggesting that this equation holds for any function? Only it doesn't hold for arbitrary $f(t)$. $\endgroup$ – Graham Hesketh Aug 5 '13 at 18:07
  • $\begingroup$ The question seems somewhat incomplete. $\endgroup$ – André Nicolas Aug 5 '13 at 18:09
  • $\begingroup$ I am sorry, I forgot to mention that f is continous and from closed [0,1] to reals. See edit above. $\endgroup$ – user191919 Aug 5 '13 at 19:24
  • $\begingroup$ If you subs $x=0$ in the equation, you will get $\frac{c}{24}=-\int_{0}^{1} t^2f(t) dt$. $\endgroup$ – Mhenni Benghorbal Aug 5 '13 at 20:52
  • $\begingroup$ My attempt at the solution (and I suppose its right): With $x=0$, I got $\frac{c}{24}=\int_{0}^{1} t^2f(t) dt $ like Mhenni Benghorbal suggested. $\endgroup$ – user191919 Aug 6 '13 at 0:53
1
$\begingroup$

Plugging in $x=0$ and $x=1$ in the desired identity, one sees that $$ 24\int_0^1t^2f(t)\,\mathrm dt=c=7-24\int_0^1f(t)\,\mathrm dt. $$ There is no reason to expect that the LHS and the RHS coincide hence, in general, there is no $c$ such that the identity holds for every $x$ in $(0,1)$ (and in fact, for $x=0$ and $x=1$).

Finding $f$ such that the identity holds is another question: assuming that $f$ is a solution and differentiating with respect to $x$ yields $f(x)=-x^2f(x)+x^7+x^5$, hence $f(x)=x^5$. This is indeed a solution and, using the LHS of the identity above, one sees that $c=24\cdot\frac18=3$.

$\endgroup$
  • $\begingroup$ Neither Taylor expansion nor integral remainder here. $\endgroup$ – Did Aug 6 '13 at 13:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.